On the dimension-free control of higher order truncated Riesz transforms by higher order Riesz transforms

Fix a positive integer $k$. Let $R_k$ be a higher order Riesz transform of order $k$ on $\mathbb{R}^d$ and let $R_k^t,$ $t>0,$ be the corresponding truncated Riesz transform. We study the relation between $\|R_k f\|_{L^p(\mathbb{R}^d)}$ and $\|R_k^t f\|_{L^p(\mathbb{R}^d)}$ for $p=1$, $p=\infty,$ and $p=2.$ We do this by analyzing the factorization operator $M_k^t$ defined by the relation $R_k^t=M_k^t R_k.$ The operator $M_k^t$ is a convolution operator associated with an $L^1$ radial kernel $b_{k,d}^t(x)=t^{-d}b_{k,d}(x/t),$ where $b_{k,d}(x):=b_{k,d}^1(x).$ We prove that $b_{k,d} \ge 0$ only for $k=1,2.$ We also show that for fixed $k\ge 3$, $$\lim_{d\to \infty}\|b_{k,d}\|_{L^1(\mathbb{R}^d)}=\infty.$$ This contrasts with the cases $k=1,2$, where it is known that $\|b_{k,d}\|_{L^1(\mathbb{R}^d)}=1$. Finally, we show that for any positive integer $k$, the Fourier transform of $b_{k,d}$ is bounded in absolute value by $1.$ This implies the contractive estimate $$\|R_k^t f\|_{L^2(\mathbb{R}^d)}\le \|R_k f\|_{L^2(\mathbb{R}^d)}$$ and an analogous estimate for general singular integrals with smooth kernels for radial input functions $f.$